AB is a vertical tower. The point A is on the ground and C is the middle point of AB . The part CB subtend an angle α at a point P on the ground. If AP = nAB , then the correct relation is

Correct answer: D. n = (2n^ 2 + 1)tan α In ∆ PAB, tan β = (AB)/(AP) In ∆ PAC, tan θ = (AC)/(AP) ∴ tan α = (β - θ) =(tan β - tan θ)/(1 + tan β tan θ) = ((AB)/(AP) - (AC)/(AP))/(1 + (AB)/(AP) ⋅ (AC)/(AP))...(i) ∵ AP = n (AB) = n (2AC) ( ∵ C is the mid point of BA ) From Eq. (i), we get tan α = (1)/(n) - (1)/(2n) 1 + 1 2n^ 2 = n 2n^ 2 + 1 ⇒ n = (2n^ 2 + 1) tan α

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Heights and Distances

NTA Abhyas 2020 · Mathematics · Heights and Distances

AB is a vertical tower. The point A is on the ground and C is the...

AB

is a vertical tower. The point

A

is on the ground and

C

is the middle point of

AB

. The part

CB

subtend an angle

\alpha

at a point

P

on the ground. If

AP = nAB

, then the correct relation is AB is a vertical tower. The point A is on the ground and C is the middle point of AB . The part CB subtend an angle α

Options

A $n = \left(n^{2} + 1\right)tan \alpha$
B $n = \left(2n^{2} - 1\right)tan \alpha$
C $n^{2} = \left(2n^{2} + 1\right)tan \alpha$
D $n = \left(2n^{2} + 1\right)tan \alpha$

Correct answer

D. $n = \left(2n^{2} + 1\right)tan \alpha$

Step-by-step solution

∆ PAB, tan \beta = \dfrac{AB}{AP}

∆ PAC, tan \theta = \dfrac{AC}{AP}

∴ tan \alpha = \tan \left(\beta - \theta\right)<br /><br />=\dfrac{tan \beta - tan \theta}{1 + tan \beta tan \theta}

= \dfrac{\dfrac{AB}{AP} - \dfrac{AC}{AP}}{1 + \dfrac{AB}{AP} ⋅ \dfrac{AC}{AP}}...(i)<br /><br />

∵ AP = n \left(AB\right) =

n \left(2AC\right)

(

∵ C is the mid point of

)<br /><br />From Eq. (i), we get<br /><br />

tan \alpha = \dfrac{\dfrac{1}{n} - \dfrac{1}{2n}}{1 + \dfrac{1}{2n^{2}}} = \dfrac{n}{2n^{2} + 1}

<br /><br />

⇒ n = \left(2n^{2} + 1\right)$ tan \alpha

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